# Chapter 11: Properties of Gases¶

## Exa 11.1¶

In [ ]:
#A certain piece of apparatus of constant volume is filled with nitrogen
#at 15 psia, 80 F. from a nitrogen bottle on a weighing scale exactly 3 lb
#of nitrogen is added to the apparatus. The final pressure and temperature
#are 25 psia, 75 F. Find the volume of the apparatus.
import math
#initialisation of variables
P= 15. 								#psia
T= 80. 								#F
m= 3. 								#lb
P1= 25.								#psia
T1= 75.								#F
#CALCULATIONS
r= (P*(460+T1))/(P1*(T+460))		#ratio
m2= m/(1-r) 						#Mass 2
V2= (m2*55.16*(460+T1))/(P1*144.) 	#Volume 2
#RESULTS
print '%s %.2f' %('Volume of the apparatus (cu ft) = ',V2)
raw_input('press enter key to exit')

Volume of the apparatus (cu ft) =  60.64


## Exa 11.2¶

In :
#Given that O2 has a gas constant of 48.3 and is a diatomic gas, compare its
#actual Cp at 100, 500, 1500 F with the values computed from the simple KTG
#initialisation of variables
R= 48.3 					#ft lb/lb R
T= 100 						#F
T1= 500 					#F
T2= 1500 					#F
k= 1.4
k1= 1.36
k2= 1.31
#CALCULATIONS
dc= R/778. 					#Cp-Cv
cp= (k/(k-1))*dc 			#Specific heat at constant pressure
cv= cp/k 					#Specific heat at constant volume
cp1= (k1/(k1-1))*dc 		#Specific heat at constant pressure 2
cv1= cp/k1 					#Specific heat at constant volume 2
cp2= (k2/(k2-1))*dc 		#Specific heat at constant pressure 3
cv2= cp2/k2  				#Specific heat at constant volume 3
#RESULTS
print '%s %.3f' %('#Specific heat at constant pressure (Btu/lb R) = ',cp)
print '%s %.3f' %(' \nSpecific heat at constant volume (Btu/lb R) = ',cv)
print '%s %.3f' %(' \nSpecific heat at constant pressure 2 (Btu/lb R) = ',cp1)
print '%s %.3f' %(' \nSpecific heat at constant volume 2 (Btu/lb R) = ',cv1)
print '%s %.3f' %(' \nSpecific heat at constant pressure 3 (Btu/lb R) =',cp2)
print '%s %.3f' %(' \nSpecific heat at constant volume 3 (Btu/lb R) = ',cv2)
raw_input('press enter key to exit')

#Specific heat at constant pressure (Btu/lb R) =  0.217

Specific heat at constant volume (Btu/lb R) =  0.155

Specific heat at constant pressure 2 (Btu/lb R) =  0.235

Specific heat at constant volume 2 (Btu/lb R) =  0.160

Specific heat at constant pressure 3 (Btu/lb R) = 0.262

Specific heat at constant volume 3 (Btu/lb R) =  0.200
press enter key to exit

Out:
''

## Exa 11.4¶

In :
#Nitrogen in a frictionless adiabatic process, expands from an initial state
#of 100 psia, 140 F to a final pressure of 10 psia. How much does the enthalpy
#change?
import math
#initialisation of variables
P= 10. 						#psia
P1= 100. 					#psia
T= 140. 					#F
k= 1.4
R= 55.16 					#ft lb/lb R
#CALCULATIONS
dh= (k*R*(T+460)/(k-1))*(math.pow((P/P1),((k-1)/k))-1)*(72./56000.)
#RESULTS
print '%s %.2f' %('Enthalpy change (Btu/lb) = ',dh)
raw_input('press enter key to exit')

Enthalpy change (Btu/lb) =  -71.79
press enter key to exit

Out:
''

## Exa 11.5¶

In :
#Air initially at 100 psia, 2000 F, expands reversibly and adiabatically
#to 15 psia. Find the change of enthalpy and of V by perfect gas laws, and
#by variable specific heats using the gas tables
import math
#initialisation of variables
P= 100 											#psia
P1= 15 											#psia
T= 2000 										#F
k= 1.4
R= 53.34 										#ft lb/lb R
cp= 0.240 										#Btu/lb R
#CALCULATIONS
v1= (R*(T+460)/(P*144))*math.pow((P/P1),(1/k))	#Initial volume
T1= (T+460)*(P1*v1/(P*(R*(T+460)/(P*144)))) 	#Initial temperature
dh= cp*(T1-T) 									#Change in enthalpy
dv= v1-(R*(T+460)/(P*144)) 						#Change in volume
print('case 1')
print '%s %.2f' %('\n change in volume =', dv)
print('\n case 2')
T2=1500 										#F
v2=R*(T+460)/(P*144)/0.241						#Volume in case 2
T2=(T2+460)*(P1*v2/(P*(R*(T2+460)/(P*144)))) 	#Temperature in case 2
deltah=0.276*(T2-460-T) 						#Change in enthalpy
dv2=v2-(R*(T+460)/(P*144)) 						#Change in volume
print '%s %.2f' %('\n change in volume (cu ft/lb) = ', dv2)
print('At T1=2460 R, from table 1, case 3')
h1=634.34
pr1=407.3
pr2=pr1*P1/P 									#pressure 2
T2=1075 										#F
h2=378.44
deltah=h2-h1 									#Change in enthalpy
v2=53.34*(T2+460)/(P1*144) 						#Final volume
dv3=v2-(R*(T+460)/(P*144)) 						#Change in volume
print '%s %.2f' %('\n change in volume (cu ft/lb) = ',dv3)
raw_input('press enter key to exit')

case 1

change in volume = 23.66

case 2

change in volume (cu ft/lb) =  28.70
At T1=2460 R, from table 1, case 3

change in volume (cu ft/lb) =  28.79
press enter key to exit

Out:
''